Multiple Choice
1.(6 pts.) If f (x) = x3 and g(x) = x − 1, Which of the following shows the graph of
f (g(x))?
Solution: Since f (g(x)) = f (x − 1) = (x − 1)3 then the graph of f (g(x)) is the graph of
x3 displaced one unit to the right. The graph is shown below:
2.(6 pts.) Find the limit
√
lim
x→1+
√
x2 + 1 − 2x
.
(x − 1)2
Solution: Multiply by the conjugate of the numerator and solve:
√
√
x2 + 1 − 2x
lim
x→1+
(x − 1)2
√ !
√
x2 + 1 − 2x
x2 + 1 + 2x
√
√
√
√
= lim+
x→1 (x − 1)2 ( x2 + 1 +
x2 + 1 + 2x
2x)
(x − 1)2
√
√
= lim+
x→1 (x − 1)2 ( x2 + 1 +
2x)
1
1
√
= √ .
= lim+ √
2
x→1 ( x + 1 +
2x)
2 2
3.(6 pts.) Assume that f (x) is a continuous function which takes the following values:
x
−1
0
1 2
f (x) −10 10 −1 3
2
Which of the following conclusions can we make by using the Intermediate Value Theorem:
Solution: We see that f (−1) < 0 < f (0) and hence there is x1 ∈ (−1, 0) such that
f (x1 ) = 0 by the Intermediate Value Theorem (IVT) since f is continuous. Further
f (0) > 0 > f (1) and hence there is x2 ∈ (0, 1) such that f (x2 ) = 0 by IVT. Lastly
f (1) < 0 < f (2) and hence there is x3 ∈ (1, 2) such that f (x3 ) = 0 by IVT. We can
conclude that there are at least 3 three solutions to f (x) = 0 (note that there could be
more).
4.(6 pts.) Consider the following rational function:
x2 − 1
(with domain {x|x 6= 1, x 6= 2}).
x2 − 3x + 2
Which of the following statements is true?
f (x) =
Solution: The points of discontinuity occur at x = 1 and x = 2; the function is not
defined at these points. Recall that a function has a removable discontinuity at a point
x0 if the limit of the function at x0 exists, but does not equal the value of the function
at x0 (this could be because the function is not defined at x0 ).
So we check the limits as x approaches 1 and 2:
(x − 1)(x + 1)
x+1
2
= lim
=
= −2.
x→1 (x − 2)(x − 1)
x→1 x − 2
−1
lim f (x) = lim
x→1
However, as x approaches 2 we get:
lim+ f (x) = lim+
x→2
x→2
(x − 1)(x + 1)
x+1
= lim+
= +∞
(x − 2)(x − 1) x→2 x − 2
(x − 1)(x + 1)
x+1
= lim−
= −∞
x→2
x→2 (x − 2)(x − 1)
x→2 x − 2
So the limit of f (x) as x approaches 2 does not exist. Thus f (x) has a removable
discontinuity at x = 1 only.
lim− f (x) = lim−
cos x
5.(6 pts.) Find the derivative of f (x) = √
.
x+1
3
Solution: We use the quotient rule to compute the derivative of f (x). Note
√
√
√
(− sin x)( x + 1) − (cos x)
0
0
(cos x) ( x + 1) − (cos x)( x + 1)
√
√
f 0 (x) =
=
( x + 1)2
( x + 1)2
√
√
√ √
(2 x)(− sin x)( x + 1) − (cos x)
−2 x( x + 1) sin x − cos x
√ √
√ √
=
=
.
(2 x)( x + 1)2
(2 x)( x + 1)2
1
√
2 x
6.(6 pts.) Find the equation of the tangent line to
√
1
y = 3( 3 x + √
)
3
x
at the point (8,
(Note:
15
).
2
√
3
x = x1/3 )
Solution: First rewrite it as y = 3x1/3 + 3x−1/3 . Applying the power rule, y 0 = x−2/3 −
x−4/3 . Then
1
1
3
y 0 (8) = 8−2/3 − 8−4/3 = −
= .
4 16
16
15
3
Then the equation of the tangent line is y −
= (x − 8).
2
16
7.(6 pts.) Find the derivative of f (x) =
p
sin(x3 ).
√
x, h(x) = sin(x), and i(x) = x3 , then by the chain rule we see
1
that h(i(x))0 = h0 (i(x))i0 (x) = − cos(x3 )3x2 , and we know that g 0 (x) = √ . Using the
2 x
chain rule again we see that
Solution: Let g(x) =
f 0 (x) = (g(h(i(x))))0 = g 0 (h(i(x)))(h(i(x)))0
1
−3x cos(x3 )
(−3x cos(x3 )) = p
.
= p
2 sin(x3 )
2 sin(x3 )
4
8.(6 pts.) Compute f 0 (1), if
f (x) = (x + 1)100 (x4 + 1).
Solution: Using the product rule, power rule and chain rule:
f 0 (x) = (x + 1)100 (4x3 ) + 100(x + 1)99 (x + 1)0 (x4 + 1)
= (x + 1)100 (4x3 ) + 100(x + 1)99 (1)(x4 + 1).
Plugging in for x = 1:
f 0 (1) = 2100 (4) + (100)(299 )(2) = 2100 (4) + 2100 (100) = 2100 (104).
9.(6 pts.) If f (x) =
1
, find f 00 (x).
1+x
Solution: We can express f (x) as f (x) = (1 + x)−1 . We compute f 0 (x) using the power
rule, and chain rule.
f 0 (x) = (−1)(1 + x)−1−1 (1 + x)0 = −(1 + x)−2 .
Using the power rule again, we obtain
f 00 (x) = −(−2)(1 + x)−2−1 (1 + x)0 = 2(1 + x)−3 =
2
.
(1 + x)3
10.(6 pts.) Compute
x sin(2x)
.
x→0 1 − cos x
lim
Solution: We will solve this limit by using the following facts:
sin(x)
=1
and
sin2 (x) + cos2 (x) = 1
x→0
x
Multiply by the conjugate of the denominator
x sin(2x) 1 + cos(x)
x sin(2x)(1 + cos(x))
x sin(2x)
lim
= lim
= lim
x→0 1 − cos(x)
x→0 1 − cos(x)
x→0
1 + cos(x)
1 − cos2 (x)
x sin(2x)(1 + cos(x))
= lim
.
x→0
sin2 (x)
lim
5
Now we can arrange the product more conveniently as:
x
1
1 + cos(x)
sin(2x)
·
·
·
x→0
1
sin(x) sin(x)
1
Now we introduce 2x into the numerator and denominator we get
lim
sin(2x)
x
x
1 + cos(x)
·
·
·
x→0
2x
sin(x) sin(x)
1
Using the identities above we compute
lim 2
= 2 · 1 · 1 · 1 · (1 + cos(0)) = 4.
6
Partial Credit
You must show your work on the partial credit problems to receive credit!
11.(10 pts.) Find the equations of all tangent lines to the graph of
f (x) =
x3
+ 2x2 + 4x + 1
3
which are parallel to the line y = x + 2.
Solution: Remember that lines are parallel if they have the same slope, so in the first
place we are looking for x such that the tangent to f at x has slope equal to the slope of
x + 2, that is, equal to 1. We are therefore trying to solve the equation f 0 (x) = 1, so we
see that
f 0 (x) = x2 + 4x + 4 = (x + 2)2 ,
so f 0 (x) = 1 implies that x = −3 or x = −1. We will now find the formula for the
tangent lines at these points. Remember we have the formula y − f (x0 ) = f 0 (x0 )(x − x0 )
where x0 is the point where we want to find the formula for the tangent for f . Remember
we are interested in points where f 0 (x0 ) = 1
(−1)3
x0 = −1: then we get y−f (−1) = 1(x−−1) so we get y−(
+2(−1)2 +4(−1)+1) =
3
4
1
x + 1, simplifying gives us y + = x + 1, so isolating y gives us y = x − as a formula
3
3
for the tangent.
x0 = −3: Then we get y − f (−3) = 1(x − −3), so we get y + 2 = x + 3, isolating y
then yields us y = x + 1 as a formula for the tangent.
12.(10 pts.) Find the derivative of
f (x) = √
using the limit definition of the derivative.
7
1
x−1
Solution:
1
1
−√
f (x + h) − f (x)
x−1
x+h−1
= lim
f 0 (x) = lim
h→0
h→0
h
h
1
1
√
−√
x−1
x+h−1
= lim
·
h→0
h
√
1
1
!
√
+√
x−1
x+h−1
1
1
√
+√
x−1
x+h−1
1
1
−
x+h−1 x−1
= lim
h→0
1
1
h· √
+√
x−1
x+h−1
x − 1 − (x + h − 1)
(x + h − 1)(x − 1)
= lim
h→0
1
1
h· √
+√
x−1
x+h−1
x−1−x−h+1
(x + h − 1)(x − 1)
= lim
h→0
1
1
h· √
+√
x−1
x+h−1
−h
= lim
h→0
1
1
h(x + h − 1)(x − 1) √
+√
x−1
x+h−1
−1
= lim
h→0
1
1
(x + h − 1)(x − 1) √
+√
x−1
x+h−1
−1
=
1
1
(x − 1)(x − 1) √
+√
x−1
x−1
−1
=
2
(x − 1)2 ( √
)
x−1
−1
=
2(x − 1)(3/2)
−1
= p
.
2 (x − 1)3
8
13.(10 pts.) (a) Use the Squeeze/Sandwich Theorem to find
1 .
lim x2 cos 5
x→0
x +1
Justify each step in your argument.
Solution: Before using the Squeeze/Sandwich Theorem we first notice that
1
≤ 1.
−1 ≤ cos
x5 + 1
Since x2 is always non-negative, we can multiply by x2 both sides of the equation without
changing the inequalities,
1
2
2
−x ≤ x cos
≤ x2 .
x5 + 1
2
Since limx→0 −x
and limx→0 x2 = 0 then by the Squeeze/Sandwich Theorem we get
= 0
1
limx→0 x2 cos
= 0.
5
x +1
(b) Consider the piecewise defined function p(x) below:

1 2


x
cos
if x < 0


x5 + 1
p(x) = k
if x = 0



 sin(x)
if x > 0.
2x
Is it possible to find a value for k which makes the function p(x) continuous at x = 0?
Justify your answer for credit.
Solution: In order for p(x) to be continuous at x = 0 we need
lim p(x) = lim+ p(x) = p(0) = k.
x→0
1
2
By part (a) we have limx→0− p(x) = limx→0− x cos
= 0.
x5 + 1
sin(x)
1
sin(x)
1
lim+ p(x) = lim+
=
lim−
= .
x→0
x→0
2x
2 x→0
x
2
x→0−
Since limx→0− p(x) 6= limx→0+ p(x), then the function is discontinuous at x = 0 (independent of the value of k). Thus, no value for k will make p(x) continuous at x = 1.
9
14.(10 pts.) An object at the end of a spring has a position function given by
s(t) = 3 sin(2πt)
at time t (where time is measured in seconds (s) and distance in centimeters (cm)).
(a) What is the average velocity of the object over the time period from t = 0 to t = 1/8
seconds?
√
√
s(1/8) − s(0)
π
2
average veclocity =
= 8( ) − sin(0)) = 8 ·
= 4 2.
1/8 − 0
4
2
(b) What is the instantaneous velocity of the object at time t =
1
seconds?
8
√
1
π
v(t) = s0 (t) = 6π cos(2πt), so s0 ( ) = 6π cos( ) = 3 2π
8
4
1
seconds?
8
√
1
a(t) = s00 (t) = −12π 2 sin(2πt), so s00 ( ) = −6 2π 2
8
(c) What is the acceleration of the object at time t =
10